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Because the discrete formula for RMS, $\displaystyle X_{RMS}=\sqrt{{1 \over N}(x[1]^2+x[2]^2+...+x[N]^2)}$, is almost the same as the formula for standard deviation (assuming mean zero), except for a factor of $\sqrt{\frac{N}{N-1}}$ the RMS and standard deviation are often said to be approximately the same.

See, for example, https://www.allaboutcircuits.com/technical-articles/how-standard-deviations-relates-rms-values/

But if we use the continuous formula for RMS, $\displaystyle X_{RMS}=\sqrt{{1 \over {T_2-T_1}}\int_{T_1}^{T_2}x(t)^2 dt}$ , it wouldn't match the continuous formula for standard deviation of a continuous distribution, because it would have included the pdf $f_X(x)$ inside the integral. I.e. $\sigma_X=\sqrt{\int x^2 f_X(x)dx}$.

Is there a way to draw this comparison in the the continuous regime? Thank you.

Homero Esmeraldo
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  • Your formula for the discrete case is not compatible with the sample standard deviation, unless you assume the sample average is zero – Firebug Jan 24 '23 at 10:44
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    Also, CrossValidated supports equation formatting with MathJax, consider using that instead of adding figures for equations – Firebug Jan 24 '23 at 10:45
  • Does this answer your question https://stats.stackexchange.com/questions/30365/why-is-expectation-the-same-as-the-arithmetic-mean ? Because it seems to be boiling down to asking why arithmetic mean approximates the expected value. – Tim Jan 24 '23 at 10:52
  • I now added the clarification that I am assuming the mean to be zero. – Homero Esmeraldo Jan 25 '23 at 05:02
  • Thanks @Tim, your comment helped me figure out the answer. – Homero Esmeraldo Jan 25 '23 at 05:35

1 Answers1

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The reason for the lack of the $f_X(x)$ in the formula, is because that term would appear only when integrating over all possible values of X. This is not the case here. We are integrating over samples of X in the given interval $[T_1,T_2]$.

We have that $Var[X]=E[(X-\mu)^2]=E[X^2]$

To estimate the variance of the signal $x(t)$ over the interval $[T_1,T_2]$ we must calculate the average of $x^2(t)$ in that interval:

$\displaystyle {1 \over {T_2-T_1}}\int_{T_1}^{T_2}x(t)^2 dt$

Then apply the square root to get the standard deviation.

Homero Esmeraldo
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